Write equations for the reaction of propanoic acid with methanol and name any organic products.
This reaction is the formation of an ester because the reactants are a carboxylic acid and an alcohol.
Propanoic acid has 3 carbons and will have the functional group COOH:
CH3CH2COOH
Methanol has 1 carbon and an OH functional group:
CH3OH
So the equation is:
CH3CH2COOH + CH3OH ---> CH3CH2COOCH3 + H2O
The organic product is methyl propanoate because there are 3 carbons before the ester group and 1 carbon after the ester group.
BH
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